Question

The inductance of a closely wound coil is such that an emf of 3.00mVis induced when the current changes at the rate of 5.00A/s. A steady current of 8.00Aproduces a magnetic flux of 40.0 mWbthrough each turn. (a) Calculate the inductance of the coil. (b) How many turns does the coil have?

Short answer

1. The inductance of the coil is$L=6\times {10}^{-4}\mathrm{H}$
2. The number of turns of the coil is$N=120$

Step-by-step solution

Given

$$\begin{gathered} \epsilon =3.00mV=0.003V \\ \frac{di}{dt}=5.00\mathrm{A}/\mathrm{s} \\ i=8.00\mathrm{A} \\ {\mathrm{\varphi }}_{\mathrm{B}}=40.0\mathrm{\mu Wb} \end{gathered}$$

Understanding the concept

By using the equation 30-35 we can find the inductance of the coil.If we pass a current iin the turns of the solenoid we can consider it as an inductor, the current produces a magnetic flux${\mathbf{\varphi }}_{\mathbf{B}}$through the central region of the inductor. The inductance of the inductor is then defined interms of that flux as by equation 30-28.

Formula:

$$\begin{gathered} \epsilon =-L\frac{di}{dt} \\ L=\frac{N{\varphi }_B}{i} \end{gathered}$$

Calculate the inductance of the coil.

By using the equation 30-35 we can find the inductance

$\epsilon =-L\frac{di}{dt}$

Rearrange as

$$\begin{gathered} L=\frac{\epsilon }{{\displaystyle \frac{di}{dt}}} \\ L=\frac{0.003}{5.00} \\ L=6\times {10}^{-4}\mathrm{H} \end{gathered}$$

(b) Calculate the number of turns of the coil.

According to equation 30-28 the inductance can be defined as

$L=\frac{N{\varphi }_B}{i}$

We have to find the number of turns so the formula rearrange as

$N=\frac{Li}{{\varphi }_B}$

By substituting the value we get,

$$\begin{gathered} N=\frac{6\times {10}^{-4}\times 8}{40\times {10}^{-6}} \\ N=120 \end{gathered}$$